Find Maximum Subarray Divide and Conquer in DSA Go - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Maximum Subarray Sum Using Divide and Conquer

What is the output of the following Go code that finds the maximum subarray sum using divide and conquer?

DSA Go

package main
import "fmt"

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

func maxCrossingSum(arr []int, left, mid, right int) int {
    sum := 0
    left_sum := -1 << 31
    for i := mid; i >= left; i-- {
        sum += arr[i]
        if sum > left_sum {
            left_sum = sum
        }
    }
    sum = 0
    right_sum := -1 << 31
    for i := mid + 1; i <= right; i++ {
        sum += arr[i]
        if sum > right_sum {
            right_sum = sum
        }
    }
    return max(left_sum+right_sum, max(left_sum, right_sum))
}

func maxSubArraySum(arr []int, left, right int) int {
    if left == right {
        return arr[left]
    }
    mid := (left + right) / 2
    left_max := maxSubArraySum(arr, left, mid)
    right_max := maxSubArraySum(arr, mid+1, right)
    cross_max := maxCrossingSum(arr, left, mid, right)
    return max(max(left_max, right_max), cross_max)
}

func main() {
    arr := []int{-2, 1, -3, 4, -1, 2, 1, -5, 4}
    result := maxSubArraySum(arr, 0, len(arr)-1)
    fmt.Println(result)
}

A 7

B 4

C 5

D 6

💡 Hint

Think about the subarray with the largest sum in the given array.

The maximum subarray is [4, -1, 2, 1] which sums to 6.

🧠 Conceptual

intermediate

Key Idea Behind Divide and Conquer for Maximum Subarray

Which statement best describes the key idea behind the divide and conquer approach to find the maximum subarray?

A Divide the array into two halves, find maximum subarray in left half, right half, and crossing the middle, then return the maximum of these three.

B Sort the array and pick the largest element as the maximum subarray sum.

C Use a sliding window to find the maximum sum subarray in one pass.

D Calculate the sum of all subarrays and pick the maximum.

💡 Hint

Divide and conquer splits the problem into smaller parts and combines results.

The divide and conquer method splits the array, solves for left, right, and crossing subarrays, then combines results to find the maximum.

🔧 Debug

advanced

Identify the Bug in maxCrossingSum Function

What error will occur if the initial values of left_sum and right_sum in maxCrossingSum are set to 0 instead of a very small number?

DSA Go

func maxCrossingSum(arr []int, left, mid, right int) int {
    sum := 0
    left_sum := 0
    for i := mid; i >= left; i-- {
        sum += arr[i]
        if sum > left_sum {
            left_sum = sum
        }
    }
    sum = 0
    right_sum := 0
    for i := mid + 1; i <= right; i++ {
        sum += arr[i]
        if sum > right_sum {
            right_sum = sum
        }
    }
    return max(left_sum+right_sum, max(left_sum, right_sum))
}

A It may return incorrect maximum sum when all numbers are negative.

B It will cause a runtime panic due to index out of range.

C It will cause a compilation error due to variable redeclaration.

D It will always return zero regardless of input.

💡 Hint

Consider what happens if all elements are negative.

Initializing left_sum and right_sum to 0 causes the function to ignore negative sums, leading to wrong results if all numbers are negative.

❓ Predict Output

advanced

Output of maxSubArraySum for All Negative Array

What is the output of maxSubArraySum when the input array is all negative numbers?

DSA Go

package main
import "fmt"

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

func maxCrossingSum(arr []int, left, mid, right int) int {
    sum := 0
    left_sum := -1 << 31
    for i := mid; i >= left; i-- {
        sum += arr[i]
        if sum > left_sum {
            left_sum = sum
        }
    }
    sum = 0
    right_sum := -1 << 31
    for i := mid + 1; i <= right; i++ {
        sum += arr[i]
        if sum > right_sum {
            right_sum = sum
        }
    }
    return max(left_sum+right_sum, max(left_sum, right_sum))
}

func maxSubArraySum(arr []int, left, right int) int {
    if left == right {
        return arr[left]
    }
    mid := (left + right) / 2
    left_max := maxSubArraySum(arr, left, mid)
    right_max := maxSubArraySum(arr, mid+1, right)
    cross_max := maxCrossingSum(arr, left, mid, right)
    return max(max(left_max, right_max), cross_max)
}

func main() {
    arr := []int{-8, -3, -6, -2, -5, -4}
    result := maxSubArraySum(arr, 0, len(arr)-1)
    fmt.Println(result)
}

A -8

B -2

C 0

D -28

💡 Hint

The maximum subarray in all negative numbers is the least negative number.

The maximum subarray sum is -2, which is the largest single element in the array.

🚀 Application

expert

Time Complexity of Divide and Conquer Maximum Subarray

What is the time complexity of the divide and conquer algorithm for finding the maximum subarray sum for an array of size n?

A O(n)

B O(n^2)

C O(n log n)

D O(log n)

💡 Hint

Consider the recurrence relation T(n) = 2T(n/2) + O(n).

The algorithm divides the array into halves recursively and combines results in linear time, leading to O(n log n) complexity.

Practice

1. What is the main idea behind the Divide and Conquer approach to find the maximum subarray sum?

easy

A. Split the array into two halves, find max subarray in left, right, and crossing middle, then combine results

B. Sort the array and pick the largest element as max subarray sum

C. Use a sliding window to find the max sum subarray in one pass

D. Calculate sum of all subarrays and pick the maximum

Solution

1. Step 1: Understand divide and conquer strategy

   The array is split into two halves recursively to simplify the problem.

2. Step 2: Combine results from subproblems

   Find max subarray in left half, right half, and crossing the middle, then take the maximum of these three.

3. Final Answer:

   Split array, find max in left, right, crossing middle, combine -> Option A

4. Quick Check:

   Divide and conquer = split and combine max subarrays [OK]

Hint: Divide array, conquer halves, combine max sums [OK]

Common Mistakes:

• Thinking sorting helps find max subarray sum
• Confusing divide and conquer with sliding window
• Calculating all subarrays sums directly

2. Which of the following Go function signatures correctly represents the divide and conquer function to find maximum subarray sum in an array arr between indices low and high?

easy

A. func maxSubArray(arr []int, low int high int) int

B. func maxSubArray(arr int[], low int, high int) int

C. func maxSubArray(arr []int, low int, high int) (int, int)

D. func maxSubArray(arr []int, low, high int) int

Solution

1. Step 1: Check Go slice and parameter syntax

   In Go, slices are declared as []int, and multiple parameters of same type can be grouped.

2. Step 2: Identify correct function signature

   func maxSubArray(arr []int, low, high int) int uses correct slice type and groups low, high as int parameters, returning int sum.

3. Final Answer:

   func maxSubArray(arr []int, low, high int) int -> Option D

4. Quick Check:

   Go slice syntax and grouped params = func maxSubArray(arr []int, low, high int) int [OK]

Hint: Use []int for slices and group same-type params [OK]

Common Mistakes:

• Using int[] instead of []int for slices
• Returning multiple values when only sum needed
• Separating same-type parameters unnecessarily

3. What is the output of the following Go code snippet that uses divide and conquer to find maximum subarray sum?

package main
import "fmt"

func maxCrossingSum(arr []int, low, mid, high int) int {
    leftSum := -1 << 31
    sum := 0
    for i := mid; i >= low; i-- {
        sum += arr[i]
        if sum > leftSum {
            leftSum = sum
        }
    }
    rightSum := -1 << 31
    sum = 0
    for j := mid + 1; j <= high; j++ {
        sum += arr[j]
        if sum > rightSum {
            rightSum = sum
        }
    }
    return leftSum + rightSum
}

func maxSubArray(arr []int, low, high int) int {
    if low == high {
        return arr[low]
    }
    mid := (low + high) / 2
    leftMax := maxSubArray(arr, low, mid)
    rightMax := maxSubArray(arr, mid+1, high)
    crossMax := maxCrossingSum(arr, low, mid, high)
    if leftMax >= rightMax && leftMax >= crossMax {
        return leftMax
    } else if rightMax >= leftMax && rightMax >= crossMax {
        return rightMax
    } else {
        return crossMax
    }
}

func main() {
    arr := []int{-2, 1, -3, 4, -1, 2, 1, -5, 4}
    fmt.Println(maxSubArray(arr, 0, len(arr)-1))
}

medium

A. 5

B. 7

C. 6

D. 4

Solution

1. Step 1: Understand the input array and max subarray

   The array is [-2,1,-3,4,-1,2,1,-5,4]. The max subarray is [4,-1,2,1] with sum 6.

2. Step 2: Trace divide and conquer calls

   The function recursively finds max in left, right, and crossing middle, correctly returning 6.

3. Final Answer:

   6 -> Option C

4. Quick Check:

   Max subarray sum = 6 [OK]

Hint: Max subarray for example array is 6 [OK]

Common Mistakes:

• Choosing sum including -5 which lowers total
• Confusing max crossing sum with max overall
• Miscomputing indices for mid and crossing sums

4. Identify the bug in this Go function for finding max crossing sum and select the fix:

func maxCrossingSum(arr []int, low, mid, high int) int {
    leftSum := 0
    sum := 0
    for i := mid; i >= low; i-- {
        sum += arr[i]
        if sum > leftSum {
            leftSum = sum
        }
    }
    rightSum := 0
    sum = 0
    for j := mid + 1; j <= high; j++ {
        sum += arr[j]
        if sum > rightSum {
            rightSum = sum
        }
    }
    return leftSum + rightSum
}

medium

A. Initialize leftSum and rightSum to very small number (e.g. -1<<31) instead of 0

B. Change loops to run from low to mid and mid+1 to high

C. Return max of leftSum and rightSum instead of sum

D. Add 1 to mid when calculating leftSum loop

Solution

1. Step 1: Analyze initialization of leftSum and rightSum

   They start at 0, which fails if all numbers are negative, missing correct max sums.

2. Step 2: Fix initialization to very small number

   Set leftSum and rightSum to -1<<31 (minimum int) to correctly track max sums even if negative.

3. Final Answer:

   Initialize leftSum and rightSum to very small number (e.g. -1<<31) -> Option A

4. Quick Check:

   Proper initialization avoids missing negative max sums [OK]

Hint: Initialize max sums to very small number, not zero [OK]

Common Mistakes:

• Starting max sums at zero causes wrong results with negatives
• Changing loop directions incorrectly
• Returning sum instead of combined max sums

5. Given an array arr of integers, how would you modify the divide and conquer approach to also return the start and end indices of the maximum subarray along with its sum?

hard

A. Store indices globally and update them only when crossing sum is calculated

B. Change functions to return a struct with sum, start, and end; update recursive calls to track indices

C. Return only sum and find indices by scanning array after result

D. Use a separate loop after divide and conquer to find indices of max sum

Solution

1. Step 1: Modify function return types

   Change functions to return a struct containing sum, start index, and end index to track subarray boundaries.

2. Step 2: Update recursive calls and combine logic

   When combining left, right, and crossing results, also combine their indices accordingly to keep track of max subarray range.

3. Final Answer:

   Return struct with sum, start, end; update recursive calls to track indices -> Option B

4. Quick Check:

   Return sum and indices together for full max subarray info [OK]

Hint: Return sum plus indices in struct, update recursively [OK]

Common Mistakes:

• Using global variables causing bugs in recursion
• Finding indices after sum calculation separately
• Ignoring indices in crossing sum calculation